A particle of mass 0.5 kg is subjected to a force which varies with distance as shown in figure

1. The work done by the force during the displacement from `x=0` to `x=4and x=4` to `x=12`are respectively
`{:((1)20J","40J,(2)40J"," 40J,(3)20J","60J,(4)40J","80J):}`
2. The increaase in kinetic energy of the particle during its displac ement from `x=0` to `x=12` m is
`{:((1)20J,(2)40J,(3)80J,(4)120J):}`
3. If the speed of the particle at `x=0` is 4 m/s, its speed st `x=8` m is
`{:((1)8m//s,(2)16m//s,(3)32m//s,(4)4m//s):}`

Explanation

1. Answer (3) During the displacement `x=0` to `x=4,` work done is `W=1/2xx` vbase x height `=1/2xx4xx10`
During the displacement `x=4` to `x=12.` work done is `W=4xx10+1/2xx4xx10=60J`
2. Answer (3) During the displacement `x=0` to `x=12` work done = total area `=1/2xx4xx10+xx410+1/2xx4xx10=80J`
This is equal to increase in kinetic energy.
3 Answer (2) By work-eneergy theorem`W_(total)=DeltaKE` During the displacement `x=0` to `x=8` `W=1/2xx4xx10+4xx10=60J`
`KE=1/2xx0.5xx4^(2)=4J` By work energy theorem, `W=KE_(f)-KE_(i)60=1/2mv^(2)-4`
`1/2xx0.5xxv^(2)=64impliesv=16m//s`